# Improper integral of the square of the solution of the advection equation

Consider the advection equation $u_t + c\, u_x =0$ with initial condition $u(x,0)=f(x)$, so that $u(x,t) = f(x-c\,t)$ is the solution.

It is asked to show that $$\int_{-\infty}^{\infty} u^2(x,t)\, dx = \int_{-\infty}^{\infty} u^2(x,0)\, dx =\int_{-\infty}^{\infty} f^2(x)\, dx$$ for $t$ fixed, with the assumption that $u(x,t) \rightarrow 0$ as $\vert x \vert \rightarrow 0$.

What I have done is $$\int_{-\infty}^{\infty} u^2(x,t)\, dx = \int_{-\infty}^{\infty} f^2(x-ct)\, dx. \qquad (*)$$ Letting $y=x-ct$ it follows that $dy=dx$, and $y \rightarrow \pm \infty$ as $x \rightarrow \pm \infty$. So that $$(*) = \int_{-\infty}^{\infty} f^2(y)\, dy = \int_{-\infty}^{\infty} f^2(x)\, dx = \int_{-\infty}^{\infty} u^2(x,0)\, dx.$$ So I haven't use the assumption that $u(x,t) \rightarrow 0$ as $\vert x \vert \rightarrow 0$. Where should I use this?

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Are you sure it's not as $|x|\to \infty$? Otherwise, all it says is that $f(x)=0$ for $x\leq 0$. – Glen O Apr 15 '13 at 11:30

Multiply the advection equation by the complex conjugate of $u$ and integrate with respect to $x$:

$$\int_{-\infty}^{\infty} dx \: u^* u_t + c \int_{-\infty}^{\infty} dx \: u^* u_x = 0$$

Note that

$$[u^* u]_{t,x} = 2 \Re{[u^* u_{t,x}]}$$

so that, when we take the real part of the above equation, we get

$$\int_{-\infty}^{\infty} dx \: [u^* u]_t + c \int_{-\infty}^{\infty} dx \: [u^* u]_x = 0$$

or

$$\frac{\partial}{\partial t} \int_{-\infty}^{\infty} dx \: u^* u + c [u^* u]_{-\infty}^{\infty} = 0$$

This is where you use the fact that $u$ vanishes as $x \rightarrow \pm \infty$; this implies that the term in brackets on the right vanishes. Therefore

$$\frac{\partial}{\partial t} \int_{-\infty}^{\infty} dx \: u^* u=0$$

which means that

$$\int_{-\infty}^{\infty} dx \: u^* u$$

is independent of $t$.

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I also added an (hoping good) answer of myself where I use the vanishing property. But is still wonder what is wrong with my first deduction without using the property. – simon Apr 15 '13 at 13:34
It's not necessarily true that $y \rightarrow \infty$; you could have $x-c t$ be constant. – Ron Gordon Apr 15 '13 at 13:40
A well-deserved $20$k. Nicely done, +1. – 1015 Apr 15 '13 at 13:51
@julien: not there yet, but thanks, man. To celebrate, I'm going to have two heart-attack burgers. I will then change my avatar to that great mathematician of space odds, Jabba the Hutt. You're up next. – Ron Gordon Apr 15 '13 at 13:55
Well, there you go. Enjoy the burgers! – 1015 Apr 15 '13 at 17:49

$$\frac{\partial}{\partial t} \int_{-\infty}^{\infty}u^2dx= \int_{-\infty}^{\infty} 2\,u \, u_t\,dx$$ using $u_t=-c\,u_x$ gives $$\int_{-\infty}^{\infty} 2\,u \, u_t\,dx = -2\,c\,\int_{-\infty}^{\infty} u \, u_x\,dx$$ by integration by parts this becomes $$-2\,c\,\int_{-\infty}^{\infty} u \, u_x\,dx =-2c\left[ uu\right]_{-\infty}^{\infty} + 2c \int_{-\infty}^{\infty} u_x \, u\,dx = 2c \int_{-\infty}^{\infty} u_x \, u\,dx$$ Since $u$ vanishes at $\pm \infty$. From the last two equation we see that $$\frac{\partial}{\partial t} \int_{-\infty}^{\infty}u^2dx=0,$$ so that $\int_{-\infty}^{\infty}u^2dx$ must be independend of $t$.

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